Question

If the function $f\left(x\right)=\{\begin{array}{cc}\frac{1}{x}{\log }_e\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right),& x<0\\ k,& x=0\\ \frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+1}-1},& x>0\end{array}$
is continuous at $x=0,$ then $\frac{1}{a}+\frac{1}{b}+\frac{4}{k}$ is equal to:

• A. $5$
• B. $4$
• C. $-4$
• D. $-5$

Answer 1

The correct option is D $-5$

Given: $f\left(x\right)=\{\begin{array}{cc}\frac{1}{x}{\log }_e\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right),& x<0\\ k,& x=0\\ \frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+1}-1},& x>0\end{array}$
As the function is continuous at $x=0$, so
$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)$
Now,
$\underset{x\to 0^{-}}{lim}\frac{{\log }_e(1+\frac{x}{a})-{\log }_e(1-\frac{x}{b})}{x}=k$
Using L'Hospital's Rule
$\Rightarrow \underset{x\to 0^{-}}{lim}\frac{\frac{1}{1+\frac{x}{a}}\times \frac{1}{a}-\frac{1}{1+\frac{x}{b}}\times (-\frac{1}{b})}{1}=k\Rightarrow \frac{1}{a}+\frac{1}{b}=k\cdots \left(1\right)$
Also,
$\underset{x\to 0^{+}}{lim}\frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+1}-1}=k\Rightarrow \underset{x\to 0^{+}}{lim}\frac{-2{\sin }^{2}x}{\sqrt{x^2+1}-1}=k\Rightarrow \underset{x\to 0^{+}}{lim}\left(\frac{-2{\sin }^{2}x}{(\sqrt{x^2+1}{)}^2-1^2}\right)(\sqrt{x^2+1}+1)=k\Rightarrow \underset{x\to 0^{+}}{lim}\left(\frac{-2{\sin }^{2}x}{x^2}\right)(\sqrt{x^2+1}+1)=k\Rightarrow k=-4$
Hence,
$\frac{1}{a}+\frac{1}{b}+\frac{4}{k}=k+\frac{4}{k}\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{4}{k}=-4-1=-5$